The given P-V graph illustrates the reversible isothermal expansion...

The work done by a gas during a reversible expansion is given by the area under the $P-V$ curve, which is mathematically represented as $W = \int_{V_i}^{V_f} P \, dV$.

From the given graph, it is evident that the isotherm for temperature $T_2$ lies above the isotherm for temperature $T_1$. This means that for any given volume between $V_i$ and $V_f$, the pressure at $T_2$ is greater than the pressure at $T_1$.

Consequently, the area under the $P-V$ curve at $T_2$ is greater than the area under the curve at $T_1$.

Therefore, the magnitude of the work done at $T_2$ is greater than the magnitude of the work done at $T_1$, i.e., $W_2 > W_1$.

Answer: $W_2 > W_1$ because the area under the $P-V$ curve at $T_2$ is greater than the area under the curve at $T_1$.